1. What is the lowest degree matric Toda bracket in $\pi_\ast(S)$ that doesn't contain zero?

By ‘degree’ I mean total homotopical degree, i.e. the $\ast$ in $\pi_\ast$. By ‘matric’ I mean to exclude ordinary Toda brackets, that is matric Toda brackets all of whose entries are one-by-one matrices, and also to exclude brackets that are trivially determined by ordinary Toda brackets.

I'm also interested in the title question with ‘first’ replaced by ‘simplest’. For instance:

2. What is the lowest order matric Toda bracket in $\pi_\ast(S)$, all of whose matrix entries are sums of products of Hopf elements, that doesn't contain zero?

By ‘order’ I mean the number of entries in the bracket, i.e. whether it is 3-fold or 4-fold or 5-fold or...

Now I'd like to know the same, but with the proviso that the bracket be detectable in the classical Adams spectral sequence:

3. What is the first or simplest, interesting matric Toda bracket in $\pi_\ast(S)$ that is detectable in the $\mathrm{E}_2$ term of the classical Adams spectral sequence?

Some remarks:

I believe the matric Massey product $\langle h_2^2, h_0, \left(\begin{array}{cc} h_1 h_3 & h_2^2 \end{array} \right), \left(\begin{array}{c} h_1 \\ h_2 \end{array}\right)\rangle $ in the $\mathrm{E}_2$ term of the Adams spectral sequence is the class $e_0$, but that is not a permanent cycle.

1 Answer
1

I know this post is quite old, but in case you are still interested, or anyone else is, I thought about sharing my recent thoughts about the topic. After all, this is the second result on "matric toda bracket" on google, and it frustrated me several times that this is still unanswered.

The bracket André suggested sadly uses relations which are only valid in $tmf$: in $\pi(S)$, we have $\nu^3 + \eta \epsilon = \eta^2 \sigma \neq 0$.

We want to construct a nontrivial matric Toda bracket, so we have to use a relation which cannot be written as a single product. The first one of those is given by $4\nu + \eta^3 = 0$.
Multiplying this by two relates this in some sense to the easier relation $8\nu = 0$, so we could wonder about this enabling us to exhibit the Toda bracket $\langle \nu, 8, \nu\rangle$ as twice the matric Toda bracket
$$\left\langle \left(\begin{array}{cc} \nu & \eta\end{array}\right), \left(\begin{array}{cc} 4 & \eta^2\\ \eta^2 & 4\end{array}\right), \left(\begin{array}{c} \nu \\ \eta\end{array}\right)\right\rangle$$
Now with matric Toda brackets we almost always get mixed degrees, so we of course want to look at the homogeneous degree $7$ part of that. The indeterminacy of degree $7$ is given by $\nu \cdot \pi_4 + \pi_4 \cdot \nu + \eta \cdot\pi_6 + \pi_6 \cdot \eta = 0$.

We want to multiply the middle term by $2$. In order to make sense of that, consider the following setting: In a good model for spectra, we can think of bimodules over a ring spectrum $E$, i.e. there are maps $E \wedge X\rightarrow X$ and $X\wedge E \rightarrow X$ plus respective commutative diagrams. Then we can talk about Toda brackets $\langle a, x, b\rangle$, where $a,b\in \pi_* E$ and $x\in \pi_* X$. Now given a bimodule map $f:X \rightarrow Y$, a defining system for the Toda bracket $\langle a, x, b\rangle$ can be pushed forward using $f$ to obtain a defining system for the Toda bracket $\langle a, f_* x, b\rangle$, so we obtain
$$
f_* \langle a, x, b \rangle \subseteq \langle a, f_* x, b\rangle
$$
This statement immediately generalizes so matric Toda brackets.

Notice that we can use this here: $S\xrightarrow{2} S$ has a representative which commutes with the right and the left action of $S$ on itself, since it can be factored naturally through the pinching map $S\rightarrow S\vee S$. (Any map between spectra is a sphere bimodule map? Not sure right now.)

In fact, the right hand side splits into usual Toda brackets, and the degree $7$ part of that will be $\langle \nu, 8, \nu\rangle$.

This element is usually known as $8\cdot \sigma$, which is nonzero. Since multiplication by $2$ is injective on the $7$-stem, we see
$$
\left\langle \left(\begin{array}{cc} \nu & \eta\end{array}\right), \left(\begin{array}{cc} 4 & \eta^2\\ \eta^2 & 4\end{array}\right), \left(\begin{array}{c} \nu \\ \eta\end{array}\right)\right\rangle = 4\sigma
$$

What this example tells us is that matric Toda brackets really should be expected all over the place: Whenever a simple relation (consisting of a single product) lifts to somewhere as a more complicated relation (with more than one summand), we should be able to lift Toda brackets built from the simple relation to matric Toda brackets by adding additional rows which account for the new summands, in our case the $\eta^3$. Note also that the exact way in which we did that did not matter that much, we could also have chosen something like
$$\left\langle \left(\begin{array}{cc} \nu & \eta^2\end{array}\right), \left(\begin{array}{cc} 4 & \eta\\ \eta & 2\end{array}\right), \left(\begin{array}{c} \nu \\ \eta^2\end{array}\right)\right\rangle
$$
or similar. The only way this generally might go wrong is that you are unable to do this in a way which actually reduces back to the bracket you wanted (we used $2\cdot \eta^2$ to end up with a diagonal matrix, and happily noticed that the component coming from the lower right entry did not interfere with the degree we were interested in).

If you look closely into the reference André posted, there a similar thing happened, only that the lifting did not happen along an actual map, but from the $E^{\infty}$-page of a spectral sequence to the actual thing, because of nontrivial extensions.